Answer
Yes
No
Work Step by Step
The function $f(x)=\tan x$ is not an one-to-one function over its whole domain, therefore in order for the function to have an inverse we must restrict the domain so that the function is one-to-one.
For example the domain of $f$ might be $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$. The restriction of $f(x)$ has the domain and range:
$D_f=\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
$R_f=(-\infty,\infty)$
The inverse $f^{−1}(x)=\tan^{−1} x$ has the domain and range:
$D_{f^{−1}}=(-\infty,\infty)$
$R_{f^{−1}}=\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
We study $\tan (\tan^{-1} x)$:
$\tan^{-1} x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
$\tan(\tan^{-1} x)\in (-\infty,\infty)$
So we have:
$\tan (\tan^{-1} x)=x$ is defined for any real $x$.
We study $\tan^{-1} (\tan x)$:
$\tan x$ doesn't make sense for all values of $x$, but only for $x$ in the domain $D_f$. So we have:
$\tan^{-1} (\tan x)=x$ is not defined for any real $x$.
Example: $x=7\pi$
$\tan^{-1}(\tan 7\pi)=\tan^{-1} (\tan 0) \pi)=\tan^{-1} (0)=0\not=7\pi$
